## Monday, May 19, 2008

### Geometry Problem 96 See complete Problem 96
Similar Triangles, Incenters, Parallelogram. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

1. join B to O1 & O extend to P ( P on AC )(same bisector)
extend GO3 to Q ( Q on AC )

ang APB = 1/2 B + C
ang MQG = 1/2 G + C ( AB//MF => ang B = ang G )
=>
ang APB = ang MQG
=>
PB // QO2

ang DEA = ang C ( DE // AC )
=>1/2 DEA = 1/2 C
=>
O2O1 // CO

2. To Antonio:
The written enunciate of problem 96 is a repetition of problem 94. The figure is all right.

1. Thanks Nilton. The written enunciate of problem 96 has been updated.

3. Alternative solution to problem 96.
Points O, O1 and B are collinear, on the bisector of ang(ABC). Points O2, G and O3 are collinear, on the bisector of ang(MGC). As triangles ABC, DBE, FGE and MGC are similar,
then OB/BC = O1B/BE = O2G/EG = O3G/GC = (O1B+O2G+O3G)/(BE+EG+GC).
But BE + EG + GC = BC, so OB = O1B + O2G + O3G, which is the same as OB – O1B = O2G + O3G. Besides OO1 = OB – O1B = O2G + O3G = O2O3.
Being ABC and BGC similar, then
ang(ABO) = ang(MGO3), and, as AB // MR, then
OO1 // O2O3.
We can see now that OO1O2O3 is a parallelogram because it has two opposite sides equal and parallel.